What is the signficance and intended purpose of variable assignments in first order logic?

Question

I have what might be a strange question. In so many words: what use is the notion of a variable assignment in first-order logic? Why care about variable assignments at all?

I'm not asking "what is a variable assignment?" I know what a variable assignment is: it's a function that maps from every variable in a (fragment of) a language to an element of the domain in the model for that language.

My question is: why should I care? When will this be useful? I'm not asking to be flippant. I feel like I can't fully understand what variable assignments are util I know what they could be used for and why (if at all) they are signficant. Here's why I'm struggling to see why they are.

• First, from what I understand, a variable assignment assigns a referent to a free variable. But overwhelmingly, I'll only ever be working with with sentences -- i.e. WFFs that contain no free variables.
• Second, it seems that I'm meant to think that a model $\mathfrak{M} = <D, I> $ or structure $\mathfrak{S} = <D, R_1 ... R_n, F_1 ... F_n>$ [this gets taught in different ways] is somehow incomplete or bad if it fails to provide a variable assignment for each variable in the (fragment of the) language we're working with. But why should I think this? After all, I'll almost never encounter a free variable, so I won't be concerned with the referent of the variable at all, and the whole point of variables is to substitute them, anyways. Why would I ever want a fixed assignment for a variable, given the high likelihood I'll be doing some substitution into it anyways?

Is the idea just that on the extremely low chance I encounter a free variable on its own or in an atomic WFF I'll need to know how to interpret it? That makes some sense, but it still conflcits with my intution that we shouldn't want any interpretation of variables at all. Why should I want it to be the case that, say, x = [____] at all? I'm inclined to think it's preferable to say that variables don't have referents, and consequently free variables and formulas containing free variables just fail to refer or don't have meaning. What am I missing?

OK. Sorry for the weird question. I'm just trying to figure out exactly what the significance of variable assignments is. I must be missing something.

Answer 1

Even though a sentence like $\forall x. P(x)$ is a closed formula, you need 'variable assignment' to express what validity in a model $M$ means. By definition, $M \models \forall x. P(x)$ means that for all assignments $\eta$ of $x$, $M \models_\eta P(x)$.

(Note that this is essentially the point the lemontree is making in a comment.)

Of course, if you really wanted, you could temporarily look at a new language structure, one in which there is an additional constant symbol that you're going to use to hold the semantics of $x$, and say that $M \models \forall x P(x)$ means that $M,a \models P(x)$ for all $a$. But that's just a way of writing down the variable assignment in a different way.

Answer 2

Variable assignments are useful for talking about parameters and for talking about definable sets.

You can talk about parameters by using additional constant symbols, but definable sets are basically collections of variable assignments and I can't think of a reasonable way to talk about them without using a variable assignment or something similar.

Here's an example of a question that uses definable sets. This question is a little contrived.

Suppose we are interested in whether the function $\sin$ is definable in $(\mathbb{R}, 0, 1, +, -, *)$ in structures that satisfy $\text{Th}(\mathbb{R})$, i.e. the real closed fields.

What does it mean for a function to be definable?

It means that the graph of the function is a definable set.

So, we want to know whether the set $\{ (x, y) : y = \sin(x) \}$ is equivalent to any sets of the form $\{ (x, y) : \varphi(x, y, \vec{p}) \}$ where $\varphi$ is a formula with vocabulary $(0, 1, +, -, *)$ and $\vec{p}$ is a collection of parameters in $\mathbb{R}$.

I don't have a full proof of this fact, but we can prove that $\sin$ is not 0-definable by noting that, if it were 0-definable, then $\pi$ would be 0-definable as the solution to $\sin(x) = 0$ between $3$ and $4$ and $\pi$ is not 0-definable in real-closed fields.